Question

Gaurav takes a loan of Rs. 45,000 from the bank at the rate of 10% p.a. compound interest. The loan is repaid in 3 installments which are in arithmetic progression. The total repayment done by Gaurav is Rs. 54,000. The amount of repayment done by him in year 1?
1. Rs.18000
2. Rs.21000 
3. Rs.16500
4. Rs.19500
5.

Answer 1

Correct Answer - Option 4 : Rs.19500

CALCULATION:

As the installments are in A.P.,

∴ Let the installments paid in the year 1, year 2 and year 3 be Rs. (a - d), Rs. a and Rs. (a + d) respectively.

Total repayment done by Gaurav = Rs. 54,000

⇒ (a - d) + a + (a + d) = 54000

⇒ 3a = 54000

⇒ a = 54000/3

⇒ a = 18000

∴ Installment paid in the year 2 = a = 18000

Now, the amount taken on loan =  Rs. 45,000

∴ Amount outstanding at the end of the year 1 = 45000[1 + 10%] - Amount paid in the year 1

⇒ Amount outstanding at the end of the year 1 = 45000(1 + 0.1) - (18000 - d)

⇒ Amount outstanding at the end of the year 1 = 31500 + d

⇒ Amount outstanding at the end of the year 2 = [(31500 + d) × 1.1] - 18000

⇒ Amount outstanding at the end of the year 2 = 16650 + 1.1d

⇒ Amount outstanding at the end of the year 3 = [(16650 + 1.1d) × 1.1]

⇒ Amount outstanding at the end of the year 3 = 18315 + 1.21d

∴ Amount outstanding at the end of year 3 = Installment paid in the year 3

∴ 18315 + 1.21d = 18000 + d

⇒ 0.21d = 18000 - 18315

⇒ d = -315/0.21

⇒ d = -1500

∴ The installments are Rs. 19500, Rs. 18000 and Rs. 16500

The amount repaid by the Gaurav in the year 1 = Rs. 19500

•  The key to finding the solution to this question is to find the common difference(d) between the installments which are in A.P.
• Amount(A) at the end of 'n' years compounded at the rate of r% is given by:

          A = P[1 + (r/100)]ⁿ where P = Principal